Limit associated with function of two variables $x$ and $y$

Question

Suppose $f(x,y)=\frac{x^2+y^2}{y}$. Find $$\lim_{(x,y) \to (0,0)} f(x,y)$$

What I have attempted: Suppose
$$x=rcos\theta$$ $$y=rsin\theta$$

Then $f(x,y)$ reduces to $f(r,\theta)$ where $ f(r,\theta)=\frac{r^2(sin^2\theta + cos^2\theta)}{rsin\theta}=\frac{r}{sin\theta}$

As $(x,y) \to (0,0) $, $r \to 0$
So, $$\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{y}=\lim_{r \to 0}\frac{r}{sin\theta}$$
$sin\theta$ is bounded function. So the above limit tends to $0$.
Therefore $$\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{y}=0$$

Am I Correct ?

Answer 1

No, it is not correct. Since $\sin\theta$ can be very small, $\frac r{\sin\theta}$ can be very large.

Just consider the points of the for $\left(\sqrt y,y\right)$, with $y\geqslant0$ and note that$$\lim_{y\to0}\frac{y+y^2}y=\lim_{y\to0}1+y=1.$$Since, when $x=0$, the limit is $0$, your limit does not exist.

Answer 2

$$\frac{x^2+y^2}{y}=\frac{x^2}{y}+y$$ and $y\to 0$ however its easy to see that the limit of $\frac{x^2}{y}$ does not exist.